ON DOUBLE POLYADICS — THE LINEAR MATRIX EQUATION. 391 



whether alike or different. Hence m p is homogeneous and of degree 

 2p in the scalar elements of the given matrices which define 6. 



Furthermore m p is non-homogeneous and of degree p (in general) in 

 the scalar elements of any particular matrix. For if the p subscripts 

 which occur in a group of terms are all alike, the p prefactors and the 

 p postf actors are all alike. 



Again, each group of terms, corresponding to a particular choice of 

 p subscripts, is homogeneous in the scalar elements of each of the 

 matrices which occur in the group. The degree in any one matrix is 

 determined by the number of times its subscript has been chosen in 

 making up p subscripts. The degree in A r is the same as that in B r . 



We thus see that each group of terms formed by the above rule is 

 made up of terms essentially unlike the terms of all the other groups, 

 each group, in other words, constitutes a distinct polynomial in the 

 scalar elements of certain of the given matrices. Within the group, 

 however, various terms may be like one another, and simplifications 

 may occur, particularly when special values are assigned to some of 

 the matrices ; for example, when some of them are allowed to become 

 the idemfactor. 



19. The Equation of Extent Unity and Order Two. 



As an example, closely allied to that of Art. 10, let us take all given 

 matrices to be of the second order, and, to begin with, let h = 1, so 

 that the linear matrix equation reduces to the simple form 



AxB = C (126) 



and can be at once solved as x = A~ ] CB~\ Forming the m p by our 

 rule we have mi = AsBs which is (an + 022) (&11 + 622)- Next 

 2m 2 = A 2 s B*s - (A*)s(B*) s . 



The scalar (A 2 )s may be expressed in terms of simpler quantities by 

 using the Hamilton-Cayley equation for the matrix A. Let the 

 coefficients in this equation be As and A"s so that 



A 2 = A S A - A" S I (127) 



where / is the identical matrix of the second order. Taking the scalar 

 of both sides of the equation we have 



(A*)s = Ah - 2A" S (128) 



and a similar equation for (B-)s- Substituting in the expression for 

 2m 2 we find on simplifying 



